yet another

[Sarahisme]

i can't quite work this one out...

 

i think the first step is to say that we want to find complex root of this:

 

z^{n} = a

 

but yeah...i dunno...can someone give me a few pointers please

 

 

 

 

Sarah

[Dave]

This looks to be a generalisation of the complex roots of unity. Solving [math]z^n = 1[/math] will give you roots [math]z = e^{\frac{2\pi k}{n}}[/math] where [math]k \in \{0, \dots, n-1\}[/math] (this follows easily from de Moivre). The sum of these roots is fairly easy.

 

Hopefully this should be enough to get you started

[Sarahisme]

ok i tryed, but i think i need a less subtle hint, i just can't do it!! :(:S

[Sarahisme]

ok here is my final attempt, god i really hope this is right....

 

well what do you guys things? :S

[timo]

I do not know what your infinite sum is supposed to be. A complex number z can be written as z = a+bi = r*exp(it), r>=0 (radius), 0<=t<pi (angle), r and t being reals. Multiplication of two complex numbers in the latter representation then is z1 * z2 = r1*exp(it1) * r2*exp(it2) = r1*r2*exp(i(t1+t2)).

Therefore z^n = r^n * exp(int). And the n-th root of a complex number is all complex numbers z' = r' exp(it') for which r'^n = r (only one solution due to restrictions r'>=0, real) and exp(int')=exp(it) (n solutions due to periodicity of exp(ix) and restriction 0<=t'<2pi).

The angle-part is the one dave mentioned (|z|=1 <=> r=1).

EDIT: No, it isn´t. I didn´t read carefully enough. Dave mentioned z=1, not |z| = 1. You have to add a "t/n" summand to the phase in Dave´s solutions to go from z=1 to |z|=1. Nevertheless, the point is: Find n angles t' for which n*t' mod 2pi = t.

 

Sidenote: The sum of all n-th roots of a complex number must be equal to zero for n>1 due to symmetry arguments, anyways.

[Sarahisme]

sorry i don't quite understand

[DQW]

You have z = a^{1/n}, where 'a' is some complex number. What is the most general polar representation for any complex number ? Use this representation for 'a'. What do you get ?

[Sarahisme]

i get

[DQW]

1. There's a small mistake there. Why don't you write it out step by step ? And can you write it in the exponential form ?

 

2. Does z change, if I replace [imath]\theta[/imath] with [imath] \theta + 2 \pi m [/imath], where m is any integer ? So, in the most general case, the argument should contain the latter, rather than the former.

[Sarahisme]

ok this is what i have done so far now...

[Sarahisme]

then do i do something like this????

[DQW]

Looks good so far. Any ideas on how to sum that series ? Does it look familiar ?

[matt grime]

Have you considered usign latex, sarah. It will help in the long run for all the typesetting you're doing and it is quite straightforward to become proficient to such a level to write the things you need to write.

[Sarahisme]

yeah i know, i have been meaning too, i will get onto it soon

[Sarahisme]

Looks good so far. Any ideas on how to sum that series ? Does it look familiar ?

 

 

well i would start by seeing what it looks like when i expand it a bit.. but then hmmmm...not sure

 

p.s. don't worry i will get the latex thing happening in the next few days

[DQW]

Isn't [imath]e^{i \theta/ n} [/imath] a constant number ?

[Sarahisme]

yeah thats what i thought, that is all the cos(theta) and sin(theta) terms will be the same, but then, i dunno doesnt that just leave you with ncos(theta) + nsin(theta)?

[Sarahisme]

dammit i didnt see what you wrote then, latex isnt showing up on my computer at the moment

[DQW]

I don't understand what post 17 is saying, but here's post #16 without the LaTeX :

 

"Isn't exp{i*theta /n} a constant number ?"

 

So can you not pull it out of the summation ?

[Sarahisme]

oh ok it makes more sense, now that i can see what you wrote with the latex in post #16

 

sorry i gotta go for about an hour, i'll be back then

[Sarahisme]

hey i'm back

[Sarahisme]

ok so here is where i am up to

[DQW]

Your first line : "Because ..." is incorrect. There is no reason for those things to be equal. This would mean that all the roots were the same and hence their sum would simply be n times the value of any one root. Also, it has no bearing on what follows.

 

Your last 2 lines are also incorrect. How are you left with a "1" in the sum after taking out exp(i*theta/n) ? What happened to all the terms ? I suggest you continue from the previous line, and procedd without skipping steps. Write each exponential term as a product of two terms - the constant exponential and the varying one. Then take the constant outside the sum .... etc.

[Sarahisme]

oops sorry about that, i should check my work more carefully, hang on i'll have another go at it!

[Sarahisme]

ok here it is: